Question

300g of mg is burned on fire to produce magnesium oxide. final product weighs 4.97. what is the empirical formula for magnesium oxide?

Answer 1

On the (guessed) data we would plump for `MgO`...

Explanation:

The empirical formula is the simplest whole number ratio defining constituent atoms in a species... And so we interrogate the moles of magnesium and oxygen in the given problem.

`"Moles of magnesium"=(300xx10^-3*g)/(24.3*g*mol^-1)=0.0123*mol`

`"Moles of oxygen"=((497-300)xx10^-3*g)/(16.0*g*mol^-1)=0.0123*mol`

And thus there are equimolar quantities of magnesium and oxygen in the given mass so we gets...an empirical formula of `MgO`..

...to do this formally we have...`Mg_((0.0123*mol)/(0.0123*mol))O_((0.0123*mol)/(0.0123*mol))-=MgO`

Answer 2

MgO is Magenesium Oxide. 497.5 gms

Explanation:

`2 Mg + O_2 = 2 MgO`

The molar mass of Magnesium is 24.31 gms/mol
So 300 gms will be 12.34 mols

Molar mass of MgO is 40.3 gms/mol

So 12.34 mols would be 497.5 gms